|
%I #23 Jun 24 2019 09:35:38
%S 2,4,8,12,16,32,64,76,114,128,170,256,512,1024,1824,2048,2166,4096,
%T 7752,8192,16384,16514,28896,32768,41154,59400,65536,68526,90914,
%U 131072,177714,230280,262144,276002,428064,524288,722400,781926,1048576,1299674,1414272,1546488
%N Sequence lists numbers k > 1 such that k^2 == phi(k) (mod sigma(k)), where phi = A000010 and sigma = A000203.
%C All powers of 2 (A000079) are part of the sequence.
%C Yes, proof: if k = 2^q with q >= 1, we have: k^2 = 2^(2q), phi(k) = 2^(q-1) and sigma(k) = 2^(q+1) - 1. Then, k^2 - phi(k) = 2^(q-1) * sigma(k). - _Bernard Schott_, Feb 20 2019
%C Contains 6*19^k for k >= 1. - _David A. Corneth_, May 29 2019
%H Giovanni Resta, <a href="/A324214/b324214.txt">Table of n, a(n) for n = 1..180</a> (first 80 terms from Robert Israel)
%F Solutions of k^2 mod sigma(k) = phi(k).
%e sigma(114) = 240 and 114^2 mod 240 = 36 = phi(114).
%p with(numtheory): op(select(n->n^2 mod sigma(n)=phi(n), [$1..1546488]));
%t Select[Range[10^5], Mod[#^2, DivisorSigma[1, #]] == EulerPhi[#] &] (* _Giovanni Resta_, May 29 2019 *)
%o (PARI) isok(k) = (k > 1) && ((k^2 % sigma(k)) == eulerphi(k)); \\ _Michel Marcus_, Feb 20 2019
%Y Cf. A000010, A000079, A000203, A324215, A324216.
%K nonn
%O 1,1
%A _Paolo P. Lava_, Feb 18 2019
|
